Does this condition on an uncountable collection of open sets have a name? Let $X$ be a topological space and $\{U_\alpha: \alpha \in \omega_1\}$ be any collection of open sets of $X$. 

The condition is this: if $\bigcap U_\alpha\not= \emptyset$, then $\bigcap U_\alpha$ is closed.

Does this condition on an uncountable collection of open sets have a name?
If I ask more, which topological property could imply this condition?
Thanks very much.
 A: The statement of the result and the proof have been corrected (I hope).
Theorem. If $X$ is a $T_1$ space with this property, then either $X$ has no infinite subset that is the intersection of $\omega_1$ distinct open sets, or $X$ is has the discrete topology. In particular, $X$ is discrete if $|X|\ge\omega_1$.
Proof. Let $X$ be a $T_1$ space with this property. Suppose that $\mathscr{U}=\{U_\xi:\xi<\omega_1\}$ is a family of distinct open sets with infinite intersection $A$. Let $F$ be any finite subset of $A$; then $U_\xi\setminus F$ is open for each $\xi<\omega_1$, so $A\setminus F=\bigcap_{\xi<\omega_1}(U_\xi\setminus F)$ is closed. Thus, $F$ is open in the relative topology on $A$, which must therefore be discrete. 
Now let $W=X\setminus A$. Let $\{A_\xi:\xi<\omega_1\}$ be a family of distinct subsets of $A$ with empty intersection, for $\xi<\omega_1$ let $V_\xi=A_\xi\cup W$, and let $\mathscr{V}=\{V_\xi:\xi<\omega_1\}$; $\mathscr{V}$ is a family of $\omega_1$ distinct open sets, so $W=\bigcap\mathscr{V}$ is closed, and $A$ and $W$ are clopen in $X$.
Finally, let $U$ be any open subset of $W$. The argument of the preceding paragraph can be applied to $\{A_\xi\cup U:\xi<\omega_1\}$ to show that $U$ is closed in $X$. Thus, every open subset of $W$ is clopen, $W$ has the discrete topology, and hence so has $X$. $\dashv$
A countably infinite set with the cofinite topology is an example of a non-discrete $T_1$ space with the property.
